r\ Physics48 (Futt 2011) Chapter23: Gauss'Law l,/ ittr-9-t.t hcY -;=- "The only thing in life that is achievedwithout efort isfailure. " - Sourceunknown "Weare what we repeatedlydo. Excellence,therefore,is not an act, but a habit." * Aristotle "Acr as if whatyou do makesa dffirence, because it does." - Sourceunknown Reading: pages605- 620 Outline: = + + = + electric flux Gauss'Law Gauss'Law andCoulomb'sLaw conductorsin electrostaticequilibrium applicationsof Gauss'Law cylindrical slmmetry planar symmetry sphericalsymmetry ProblemSolvingTechniques ' You shouldknow how to calculatethe electricflux tfuough a given surface.A useful fact to rememberis: if the field is uniform, thenthe total flux througha closedsurfaceis zero.Sometimes you can easilycalculatethe flux ofa uniform fieid throughpart ofa closedsurfacebut you have beenaskedfor the flux throughthe ranaining part. Sincethe two contributionsto the total flux sum to zero,they arethe negativesof eachother. Someproblemsdeal with Gauss'law in the form ae(D= quno, where(Dis the electricflux through a q"no closedsurfaceand is the chargeenclosedby the surface.You might be giventhe chargeand askedfor the flux or you might be given the flux (or the informationto calculateit) and askedfor the enclosedcharge. Whena problemdealswith a conductor,you shouldrememberthat the electricfield inside the conductoris 0, that the field just outsidethe conductoris perpendicularto the surfaceand has magnitudeo/e6,whereo is the areachargedensity,andthat the total chargeenclosedby any Gaussiansurfacecompletelyinsidethe conductoris 0. Manyproblemsaskyou to useGauss'lawtosolvefor the electricfield in givensituations.Become adeptin finding an expressionfor the total flux througha Gaussiansurfacefor the threespecialcases discussedin the text: cylindrical, spherical,andplanar symmetry.For a cylindrical Gaussiansurface, ^ t^ ,. with a radial field that is uniform over the curvedportion of the sr:rFace,I E . dA = (2trL)E , where r surfacewith a is the radiusofthe Gaussiancylinderand L is its length.For a sphericalGaussian of the Gaussian radiat field that is uniform over the swface,I n . d.l = (4nr2 )E , whercr is the radius chargedalsity o sphere.Also rememberthat the electricfield dueto a largeplanewith unifom area is c/2to. Mathenatical Skills Electric Flux Calculations: areaintegrals'If To calculatethe electricflux, you shouldbe familiar with the evaluationof simple to i"t"gration is a rectanglewith sidesa and b, you will probablywant position the tir","gi"r the axes'The .oordirrut "rsystemsothat the rcc1'n;le is in the a .yplanewith two of its sidesalong flux integral dy. The infinitesimalelementof areacanbelaken to be a rectanglewith sidesdx and thenbecomes drdy E,(,.vr I:=,1;=, ofthe electric Integrationsinx and] ale carriedout independently.If, for example,the z component fieldis givenby E"=9t'Y+2Y.+hen 1b /^? * = Jloli=,pf lt + zy)a"ay= Jy+lzf f I tlo + ^2ry)l}=odv = (o3y2 + *\lr=r= o'b'+ ob', = I!,=01u"'r' + ?ay)dy y' wherethe integtationoverx wascarriedout first, followed by the integrationover For manyproblemsof this chaptera Gaussianswfacecanbe chosenso that the normal on the fi rft" electricfield hai the samevalueat all points on a portion of it andis zero --p"*"t = region of the othe. portions.The integralfor the flux tlen reducesto @ EA, whercA is the area the normal of magnitude the E is and ou", *hi"h th" ,rormalcomponentof the field is not zero component, The areaofa To evaluatethe flux, you mustknow how to calculatethe areasofvarious surfaces' zR2;the surfacearea rectanglewith sidesof lengtha alldb is ab;the areaof a circle with radius-Ris with of a spiere with radiusR is 4rRz; andthes'rface meaof the rormdedportion of a cylinder radiusR andlengthI is 2rRL. imaginea If you havetrouble rememberingthatLltL givestheareaof a cylindrical surface, is the same puprito*"i roU ','.uppedexactlyoice aroundwith a towel. The surfaceareaofthe roll L' its areais lritr" *"u of *t" towel. Sincethe towel is a rectanglewith sidesof lengSh'2r nd 2mL. Calculationsof charge: You must alsobe able to calculatethe chargeenclosedby a Gaussiansurfacewhen the volume,area" or linear chmgedensityis given. If an object hasa uniform volume chargedensityp, for example, the chargeenclosedis $venby pV, where Zis the volumeof the part of the object that lies within the Gaussiansurface.Ifthe Gaussiansurfaceis completelywithin the object andthe object doesnot haveany cavities,then the regionenclosedby the Gaussiansurfaceis completelyfi]led with charge and lzis the volume enclosedby the Gaussiansurface.If the object hasa cavity that is wholly within the Gaussiansurface,then Zis the volume enclosedby the Gaussiansurfaceminus the volume of the ca\tity. _ suppose,for example,a sphereof radiusR hasa uniform volume chargedensityp andthe Gaussiansurfaceis a concentricsphereof radiusr.If r < R, then the Gaussiansphereis filled with chargeandthechargeenclosedis 4npr3/3.If,on theotherhand..R< r, thenthe Gaussiansurfaceis only partially filled with charge.The chargeenclosedis the total charge,or 4zpR3/3. Ifa sphericalobject hasa sphericalcavity with radius.R"andthe Gaussiansurfaceis within the objectbut outsidethe cavity, thenthe chargeenclosedis l(4r23 /3) - (42Rc3 B\. The first term in the bracketsis the volume enclosedby the Gaussiansurfaceandthe secondis the volume of the cavitv. IfR < r, the chargeenclosedis p[(4tn3B7 - 14rn"3/311. Now, considera solid cylinderwith radiusR andlengthz, having a uniform volr,me ch.arge densityp. Supposethe Gaussiansurfaceis a concentriccylinder with radiusr andthe samelengthas the cylinder of charge.If r <.R,the chargeenclosedis 2zqL arfi,if R < r, itis 2npRL.If the rylinder hasa cavity that is inside the Gaussiansurface,its volumemust be subtractedfrom the volumesin theseexpressions, If the volumechargedensityvariesfrom point to point in the object,you must evaluatethe ,f rnte$al p dV over the volumeof that part of the object lying within the Gaussiansurface.The J most commonexampleis a spherewith a chargedensitythat variesonly with distancer from the center.Carry out the integrationby dividing the sphereinto sphericalshellswith infinitesimal thicknessdr. A typical shell extendsffom r to r + dr andhasa volume of 4zz2dr. Notice that this is the productofthe surfaceareaofthe shell andits thickness.The chargein the shell is Axflr)rz fu.If the Gaussiansurfaceis a sphereofradius r, concenhicwith the sphereof chargeand entirely within - _ lr it, the chargeenclosedisl 4z p(r) 12dr . If the Gaussiansurfaceis entirely outsidethe charge distribution,tlre chargeenclosedis lR 4tt p1r)r'1dr . Notice the upperlimits of integrationare different for thesetwo cases. For a cylinderwitl a chmgedensitythat dependsonly on the distancer from the axis, divide the cylinder into concentriccylindrical shellswith thicknessdr. The volumeof a shell isZmL dr, wherc Z is the lengthofthe cylinder.If the Gaussiansurfaceis a concentriccylinder with mdius r andis insidetlre chargedistribution, thenthe chargeenclosedisJ" 2rrl p(r) dr. Ifthe Gaussiansurfaceis outsidethe distribution,the chargeenclosedis lR 2ttrL p(r) dr . Questions and Example Problems from Chapter 23 .ball lieswithinthehollowof a metallicsphericalball of radiusR. Forthreesituations, fliililJ*r"o are(1)+4q,0;Q) -4q, +10q; (3)+16q, -12q.Rank ontheballandshell,respectiveTy, thenetcharges to thechargeon (a)theinnersurfaceof theshelland(b) theoutersurface,most thesituationsaccording oositivefirst. ( r) Z,^\ (3) (A) (@)) \/ f,^,fn-q -aArAQ - 4 o' D ,c/dn ,\A-sIl o,Jb -t fu-i-t '" ' (t% ,,Ln"ur.'q ' +l{0.6 d4&4Are-t -,rrrur-o,z.D.q!t : n',fy IWU'. - (6t + r.lq) (r) (al Problem I The squaresurfaceshownin the figure below measures3.2mm on eachside.It is immersedin a rmiform electricfie1dwith magnitudeE = 1800N/C. The field lines makean angleof 35owith a normal to the surface,as shown.Takethat normal to be directed"outward," asthoughthe surface were one faceofa box. Calculatethe electricflux throughthe surface. => *n o 4f:,J' -:,,.l.r*D.Ke-e$ o''\je A b\ -e e .Lr.rrirpL,ni -e-0-p-clt {vsAJ -.t ip r,tsfi^rit- u-"f,-, 'l*;. Do N/,^ F = )f ' te fi = (3.2.rlot.)' ,rr" -t &fr.rx )"t o$ n*'a'-g*'-ti U Et ,1ur-e^11)rA'" -) Q= E.A'= eAr-oO j,4 f,,^€,.{ F * e.' x @: J1,.*"?tu-r'd ; e= 115"t@25' / O= EAcaeO = 6= ( trooN/c) (2"a" lot n,)\ co':115" -/.c *1o'rln'/c, Problen 2 cubeof length2.0 m and An electricfield givenby E = 4.0i -3.O(y'z+2.Qj piercesa Gaussian positionedasshownin the figure below. (The magnitudeof E is in N/C andposition x is in m.) What is the electricflux t}rough the (a) top face,(b) bottom face,(c) left face,and (d) back face?(e) What is the net electric flux throughthe cube? 4.ot - 3.o (ya-+a.o)i E t v afr0.l9 aa ",J,AaAA It- ,ttYi}J;ot p= ( q.o rs/") i - (3.oN/c) .t- (") ly"tleip W I y'cmt* e"o]5. --, ) - A.or., E= 4,Dt - 3.ofCr.";\ l"ol1 = 4,oi - tr1 '--t --> 0=) a"tn' / ( ot A-= (aaN/,-)C-(ren,l.)j) . -'t JA',= JAj *^bPry"^ - Grr)rn = -lrN/c5{A 5(r.02-re3)-(J*1) f 5dn O= = 4"o rnr-\ot- c",oa t4 4 W -/rN/c (+.on'):[ % N'.'l \/ - /\ra^),il"r"1 t -llt fu4 Jn(-e) i..i=o Q= \ E.tr = \ lU"ot- (;'o)(vla"")t]" ( ( q,ot\ ,Jnte) ),,/ -'- -16Nm'./^a CA= Y /t' = -L/.oft^ = -4"ory.(4"D ") Problem3 A chargedparticle is suspendedat the cent€rof two concentricsphericalshellsthat arevery thin and madeofnonconductingmaterial.Figure a below showsa crosssection.Figureb givesthe net flux O througha Gaussianspherecenteredon the particle, asa function of radiusr of the sphere.The scaleof the vertical axis is set by @s=5.0 x 105Nmz/C.(A) What is the chargeof the central particle?What arethe net chargesof @) shell A and (c) shell B? _o. fue* )4r., ^' : 'Y ,R- ?0 DF j-d.u)) = nn*tL, "dn -$.n.) Q a --C' rA h a) -{-orr< ro ,; +'n l,oxp5v1^x7, ..n).gr"n.t , Q= W FC -lD = (!.oroqn,,'fd (t"zsrtoloc/N,r,')ll'??^ I $nn"l 0u b) $^ e,\< c. h ., d: -5.o"ro5N*7. ,r"*-A1r",*t- r,t'.t -i C6 A = -4"q3*lo-*L . $*rr.) (-s"oxlD5N rl/c)(t.zs"lortrc'/p*'; IUur--*'-1_.:_1--.-.._.-------T_"1 q''l'i^lo-"Cl Lc) -oyr-rIalq r l% *= probtem 4 ><I0+eG-side-fu conductoris a An isolatedconductorof arbitiry .hup"h* u o"t""lhar!e'ofTT0 cavitywithin whichis a pointchargeq : +3.0x 10' C. Whatiis thecharge(a) onthqcavitywall and (b) on the outer surfaceofthe conductor? (") -.U^t^uAt-'" tu$'-t* cAaoa-c,n Pt*'t''aa o'"ui'+a/.€ qJ* i*.,, t.*''J-'r'*'fn -,t;)s"* -S;"r"" E=o ".*}xd " Yu dn4)"r-c $,-lJ }}$ + L&crAA r*^a! rcr^J-LLr!'Ua ).A d?il)a*4 )3Y -2,lr",&^'u ( b) ^'t q 5-r. "atL$an ^ <cvnAur#t, ^1 ffw h-'n-- )r-tJt ^ ryuo ^. qE,J A'= O. s)e4rYv . -.\ b;:Y]:'. o\L n - /-r , Llen") = (++ O '*.l.F.J -.. r-. -= Q b wd.q fi,* ^Aol'tr'uz'an^t"L c-!,.oo6, Q on l3' rc,o"d,u-*, i %**& = -X t^)a{P A u = | 3 " Or ' r . ' Problems / Thefigure.belowis a sectionofa conductingrod of radiusRr : 1.3QlmmandlengthL: i 1.00m insidea thin-walled coaxial conductingcylindrical shell of radiusRz: 10.0Rrandthe (same)length L. Thenet chargeon the rod is Q, : +3.40x 10rz C; that on the shellis Qz = -2.00Qr. Whatarethe (a) magnitudeE and(b) direction (radially inward or outward)ofthe electric field at radial distance r = 2.00R2?Whatare(c) E and(d) the directionat r: 5.00Rr?Whatis the chmgeon the (e) interior and(f) exterior surfaceofthe shell? f F. lA'= t . -iyrao .oad+ E(),nrL) E (a) !>ot ,-- \.ooA"- I ] . t 1 o * 1 o ' ' c - + ( - { o . f o ^l o ' ' c ) -4,1'u^^ Io-'tc- ,,;G;;''.%^.) ( a a , r d 3 ^ ) (l l - o " . ) (b) C =5-ooAr 9. eo'r 0.u €o (+) d\tt Zo - la^ I /e\ = QrtC{2, -' -- 3.{o rlO L F= O.Irr g7 / t/ 4:t /\ '= rL -- I d' tt (o fc) E(a'.cL) - q. *'/t- d\ lr go = 5E.Jf,= *gr'" I -/ cb"n"l= ) .".,4. l S.."r = $.n.\= O r 3''iO* IO-'"cI 3.,tol lC'aC -rE rL Q-.* \ Nir( 7.f 5'lo /1m') = -Q, = * po'^I ( rr'o".) ( r..so - 3.riO ^ lD-'ac --J..1o"lo-'au <o ) Problem6 A squarerretal plate of edgelength 8.0 cm andnegligiblethiclinesshasa total chargeof 6.0 x10'6C. (a) Estimatethe magnitudeE ofthe electricfield just offthe centerof the plate (al say,a distanceof 0.50mm) by assumingthat the chargeis spreadunifomrly over the two facesof the plate. h,t E = V€.ulq-p"a =) -a,vnxr ;)t}o J'^^4t /- = rfl w/r t E- = Co A = -4^r4\a4! uJ'^nZu ds"r"'*'d 2f V* t:"*$*'q. ,^, -a4r'rl.aA.,u^"+zr*'l,6 o"a#v i9,, a*v$^n "o.-.qS "d (- + <'e$u""b"'y 14r,/+a/'4tz 'ttr,* 4n'* ^A* (o) J)-rld;rt+^'!d -ao , = (ry .7t trt }i*a! g's"a^€'t (ry= ; *t"o-p'.a-u , ;trv n --))" AA = * (R= ^') = F.g^,"",-/ 6.o^ ld'c A (g.o^to-a-)t(t.vs^tf'"c"A^') ili#"U?l" *tow, a smallcircularholeof radiusR = l 80crnhasbeencutintotlremicldleof an infinite flat, nonconductingsurfacetlpt basuniform chargedensityo : 4.50 pC/m". A z axis, with its origin at the hole's center,is perpendicularto the sutface.In unit vectornotation, what is the electricfield at point P at z : 2.56 cm? ^);h a /wAr-. A "1r.f ,J,a"4' > ri"rtu-*rAub*A4,0.De+ f -- 7aU ) E=9au(t:*6 L-fMT -a E^r-**t, &) ,. e_+ . A€-I zTa, (R) ( q.so^ | d''c/.*) 1 o.o15/o'"') A ( a.ts * I d'' c/*'") { {o.oasr"-)\1o,orr'.)" F:(o"^"€N/.)R >al-.aL.Xl* tPtu"* Problem 8 Two chargedconcentricsphereshaveradii of 10.0cm and 15.0cm. The chargeon the inner sphere is4.00xlOaC,andthatontheoutersphereis2.00xl0-sC.Findtheelectricfield(a)atr:12.0cm and(b) at r : 20.0cm. ().- -+ = a.rl ln nn"Ye. E" \ f4. ln=%u-tk_ E Ulxr")= s..-/€"----) E =akL 3Tf .\ (o\) = lA,ocro, =4.oorld8C $es.) 4^. ( r . 9 q x ) o q N r l 7 c a ) C e r " o olxo j e ) A- (t)" ier.,)a ( u) -Vn r = lo.o*o, ? 6c -- 6"oo,. lo ec = $ encl { . o o x 1 9 - t g + 3 . . o o x l o = 1t"=s*io+N,/rI (o.aorn)\ !'-t Problem 9 A solid nonconductingsphereofradius R = 5.60 cm hasa nonuniformchargedistributionofvolume chmgedensityp: (14.1 pclmi) r/R, wherer is the radial distancefrom the sphere'scenter.(a) What is the sphere'stotal charge?What is the magrritudeE of the electricfield at (b) r: 0, (c) r = R/2.00, and(d) r: R? . \--'l 71I'-lf-/rn p = Ar/o / \ c! = (") \ tl A=fi.lxlo Q(c)Jv %= Ll4rAR _:--- = (+) '1.7t" ld'=c (.) t/*' A- S.Lo.r'., -6 A = tJA('coJ" '" ''' t'(A7n)4nr'dc o h = rrAB' : -(l,l.r^10'"c/.)(t.(,o^tdl-,)3 -" --lFl ( b) .t p=o,(eo") f F.rtr= .-. E A (rync)= (u*)t%,) 4""ye. _-+ f (anc) '= \*U,>"BLee*r= 1"j r-- -- ---!- (d) *-. f=A , S,*r= ?.?9"10'sc 4oe- -rA r' B c ( r"d"